After searching a bit I've found that you can't use v(t)=A*sin(2pi*f*t) or the current will be zero, so,in order to calculate the voltage for the alternate current we would use the formula Vrms=V0/squareroot of 2
and Vrms=Irms*R which would be equal to V0=I0*R

So, the final equation for Vout would be: Vout=(-1-2)*2=-6v which would be the amplitude of the curve of the tension on the output of the amplifier...
I think this's right,but if not I would like to be corrected...

Staff: Mentor

I thought that v1(t) would be zero since the frequency is 0.5 and the period is 1 and if we substitute the result of v1(t) will be zero..

Um, no. I'm not following where your confusion is coming from. V1(t) is a sinusoidal source, which is driving the input of the amplifier circuit. You are asked to solve for Vout(t), which is affected by the sinusoidal input V1(t), the DC input offset of 1V, and the gain set by the resistors around the opamp circuit.

You started with the definition of V1(t):

esmeco said:

We have a sinusoidal sign with amplitude=1v and frequency=0.5

v(t)=A*sin(2pi*f*t)

f=1/T
T=2s

So to clean up the text a bit, that means [tex]V_1(t) = sin(\frac{2\pi t}{2})[/tex]

Use that value for the V1(t) that is in your equations, and solve for Vout(t). Nothing is going to eliminate that sinusoid from your final answer.

Staff: Mentor

The peak-to-peak amplitude of the AC voltage is 2 I guess...And the DC offset voltage is 1v I guess...Should we substitute the t in the above expression to determine the amplitude of the output wave?

Almost correct. The convention is to measure the DC offset to the average (middle) part of the sine wave, which in this case is -4V. Just think of what the voltage would be if the amplitude of the sine wave were 0V, and that's the DC offset part of the output voltage waveform.

So now you've answered the original problem. You've drawn the waveform as you've been asked (be sure to label the time ticks on the horizontal axis correctly, and the voltage ticks on the vertical axis), and you know that the AC amplitude is 2Vpp, and the DC offset is -4Vdc. See, that wasn't so hard, right?

Suppose we're given an AC coupled inverting amplifier (aka. active high pass), and with a DC source at the positive input to provide offset. The circuit diagram is attached. Input Vs is assumed to be some arbitrary sinusoidal input.

How do I prove that the DC source, Vref, is not amplified at the output with such a configuration?

2. My attempt:

I know that you need to use superposition here, with the DC-decoupled AC signal and the DC offset. So I can find that the transfer function for the AC input is just the gain for an inverting amplifier.

But for the DC offset, I assumed this circuit to be a DC amplifier with Vin = 0, then I did this...

(0 - Vref) / R1 = I1
(Vout - Vref) / R2 = I2
I1 = I2

Therefore, Vout = Vref * (1/R2 - 1/R1)

But from lab, I know that Vref shouldn't be have a gain. In other words, the output should just be a sinusoidal waveform with a DC offset = Vref.